Question: Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C$. What is the degree measure of $\angle ACB$?
Let $B'$ be the point at $0^\circ$ latitude and $115^\circ$ W longitude.  We see that $\angle ACB = 360^\circ - 110^\circ - 115^\circ = 135^\circ.$

[asy]
import three;
import solids;

size(200);
currentprojection = perspective(6,3,2);

triple A, B, Bp, C;

A = (Cos(110),Sin(110),0);
B = (Sin(45)*Cos(-115),Sin(45)*Sin(-115),Cos(45));
Bp = (Cos(-115),Sin(-115),0);
C = (0,0,0);

draw(surface(sphere(1)),gray(0.9),nolight);
draw((1,0,0)..(Cos(55),Sin(55),0)..(Cos(110),Sin(110),0),red);
draw((1,0,0)..(Cos(-115/2),Sin(-115/2),0)..Bp,red);
draw(Bp..(Sin((45 + 90)/2)*Cos(-115),Sin((45 + 90)/2)*Sin(-115),Cos((45 + 90)/2))..B,red);
draw((-1.2,0,0)--(1.2,0,0),Arrow3(6));
draw((0,-1.2,0)--(0,1.2,0),Arrow3(6));
draw((0,0,-1.2)--(0,0,1.2),Arrow3(6));
draw(C--A);
draw(C--B);
draw(C--Bp);

label("$x$", (1.2,0,0), SW);
label("$y$", (0,1.2,0), E);
label("$z$", (0,0,1.2), N);
label("$110^\circ$", (0.3,0.2,0), red);
label("$115^\circ$", (0.3,-0.2,0), red);
label("$45^\circ$", (-0.3,-0.5,0.1), red);

dot("$A$", A, E);
dot("$B$", B, NW);
dot("$B'$", Bp, NW);
dot("$C$", C, NE);
dot((1,0,0));
[/asy]

Let $D$ be the point diametrically opposite $A,$ let $P$ be the projection of $B$ onto the $yz$-plane, and let $Q$ be the projection of $P$ onto line $AD.$

[asy]
import three;
import solids;

size(200);
currentprojection = perspective(6,3,2);

triple A, B, Bp, C, D, P, Q;

A = (Cos(110),Sin(110),0);
B = (Sin(45)*Cos(-115),Sin(45)*Sin(-115),Cos(45));
Bp = (Cos(-115),Sin(-115),0);
C = (0,0,0);
D = -A;
P = (B.x,B.y,0);
Q = D/2;

draw(surface(sphere(1)),gray(0.9),nolight);
draw((1,0,0)..(Cos(55),Sin(55),0)..(Cos(110),Sin(110),0),red);
draw((1,0,0)..(Cos(-115/2),Sin(-115/2),0)..Bp,red);
draw(Bp..(Sin((45 + 90)/2)*Cos(-115),Sin((45 + 90)/2)*Sin(-115),Cos((45 + 90)/2))..B,red);
draw((-1.2,0,0)--(1.2,0,0),Arrow3(6));
draw((0,-1.2,0)--(0,1.2,0),Arrow3(6));
draw((0,0,-1.2)--(0,0,1.2),Arrow3(6));
draw(C--A);
draw(C--B);
draw(C--Bp);
draw(C--D);
draw(B--P);
draw(A--B);
draw(P--Q);
draw(B--Q);

label("$x$", (1.2,0,0), SW);
label("$y$", (0,1.2,0), E);
label("$z$", (0,0,1.2), N);

dot("$A$", A, E);
dot("$B$", B, NW);
dot("$B'$", Bp, NW);
dot("$C$", C, NE);
dot("$D$", D, W);
dot("$P$", P, NE);
dot("$Q$", Q, S);
dot((1,0,0));
[/asy]

Assume that the radius of the Earth is 1.  Since $\angle BCP = 45^\circ,$ $CP = \frac{1}{\sqrt{2}}.$

Since $\angle ACB' = 135^\circ,$ $\angle PCQ = 45^\circ,$ so
\[CQ = \frac{CP}{\sqrt{2}} = \frac{1}{2}.\]Since plane $BPQ$ is perpendicular to $\overline{AD},$ $\angle BQC = 90^\circ.$  And since $CB = 2 \cdot CQ,$ triangle $BCQ$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle.  In particular, $\angle BCQ = 60^\circ,$ so $\angle ACB = \boxed{120^\circ}.$